Chapter 8.2, Problem 38AYU

To determine

How far was the moon from earth when the measurement was obtained?

Answer to Problem 38AYU

  $\boxed{193.24613km}$

Explanation of Solution

Given information:

At exactly the same time, Tom and Alice measured the angle of elevation to the moon while standing exactly $300km$ apart. The angle of elevation to the moon for Tom was $49.8974°$ . See the figure to the nearest $1000km$ , how far be the moon from earth when the measurement was obtained?

  

Calculation:

Consider the figure as per statement we can see that oblique triangle formed by Tom,Alice and Moon.

Angle $B$ and $C$ are acute but angle $A$ seems to be obtuse.

Now angle $C=49.8974°$ and side $b=300km$ ....... $\left(1\right)$

By using triangle property,

  $\begin{array}{l}A+49.9312=180°\\ A=180°-49.9312\\ A=130.0688°\mathrm{.......}\left(2\right)\\ also,\\ A+B+C=180°\\ B=180°-\left(A+C\right)\\ B=180°-\left(\mathrm{130.0.688}°+49.8974°\right)\\ B=0.0338°\mathrm{.......}\left(3\right)\end{array}$

Now apply law of sines,

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \frac{\sin 130.0688°}{a}=\frac{\sin 0.0338°}{300}=\frac{\sin 49.8974°}{c}\\ a=300\times \frac{\sin \left(130.0688°\right)}{\sin \left(0.0338°\right)}\\ a=300\times \frac{0.76527}{0.0005899}\\ a=389172.281637km\end{array}$

Hence, the distance of Moon from Tom is, $\boxed{389172.281637km}$

Now,

  $\begin{array}{l}\frac{\sin 0.0338°}{300}=\frac{\sin 49.8974°}{c}\\ c=300\times \frac{\sin 49.8974°}{\sin \left(0.0338°\right)}\\ c=300\times \frac{0.76489}{0.0005899}\\ c=\boxed{388979.0355km}\end{array}$

The distance of Moon from Alice is $\boxed{388979.0355km}$ .

Now the difference between Tom to moon and Alice to moon is,

  $\begin{array}{l}a-c=389172.2816337-388979.0355\\ =193.24613km\end{array}$

Hence, the distance of moon from earth is, $\boxed{193.24613km}$